Probability question on seating arrangements

[jackbauer]

Hi people, was wondering if someone could help me with this question. n men and m women are seated at a round table with n+m chairs. find the probability that a man amongst the n men has a woman seated immediately to his right? Find also the probability that a woman in the m women has a man either side of her? I think that the total sample space of possible arrangements from combinatorics is (n+m-1)! but i don't know how to proceed from here?Anyone got any hints?
Thanks
Jack

 

[D H]

Jack, you are making too much of this problem. It's easy. Since the table is round, everyone at the table has someone to their right and someone else to their left. No nasty endpoint conditions exist. You are asked about the gender of the person sitting in an adjacent seat. Those seats are all you need to consider. No combinatorics are needed.

 

[Architeuthis Dux]

Hello Jack,

This is an interesting probability question. To solve it, we can use the concept of conditional probability. Let's start with the first part of the question.

To find the probability that a man among the n men has a woman seated immediately to his right, we need to first determine the total number of arrangements where a man is seated next to a woman. This can be done by fixing a man in one of the n chairs and then arranging the remaining n-1 men and m women in the remaining n+m-1 chairs. This can be done in (n-1+m)! ways.

Next, we need to determine the total number of possible arrangements of n men and m women in n+m-1 chairs. This can be done in (n+m-1)! ways.

Therefore, the probability of a man having a woman seated immediately to his right is (n-1+m)! / (n+m-1)!. This simplifies to (n-1)! / (n+m-1)(n-1)! which further simplifies to 1 / (n+m-1).

For the second part of the question, we can follow a similar approach. The total number of arrangements where a woman has a man on either side can be determined by fixing a woman in one of the m chairs and arranging the remaining n men and m-1 women in the remaining n+m-1 chairs. This can be done in (n+m-2)! ways.

The total number of possible arrangements of n men and m women in n+m-1 chairs is still (n+m-1)! ways.

Therefore, the probability of a woman having a man on either side is (n+m-2)! / (n+m-1)!. This simplifies to (n+m-2)! / (n+m-1)(n+m-2)! which further simplifies to 1 / (n+m-1).

I hope this helps. Good luck with your problem-solving!